Show that the system of equations has no unique solution

  • #1
chwala
Gold Member
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Homework Statement
see attached
Relevant Equations
linear algebra
1712892630952.png

solution;

1712892671919.png


@Mark44 we discussed this sometime back on the forum. The approach of using determinant or echelon form of a matrix was sufficient.

Would i be correct to use this approach,

##x+2y+3z=4x+5y+6z##
##3x+3y+3z=0##

and

##7x+8y+9z=x+2y+3z##
##6x+6y+6z=0##

Then, it follows that the two equations are dependent i.e

##6x+6y+6z= 2(3x+3y+3z)##

thus the system cannot have a unique solution.
 
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  • #2
This is a continuation of the problem;

1712896135094.png


1712896150745.png


In my approach from 1.i,
##3x+3y+3z=0##
##x+y+z=0##
##x+y+\dfrac{1}{3} (1-x-2y)=0##
##3x+3y+1-x-2y=0##
##2x+y=-1##
##y=-2x-1##
 
  • #3
chwala said:
Would i be correct to use this approach,
##x+2y+3z=4x+5y+6z##
##3x+3y+3z=0##
and
##7x+8y+9z=x+2y+3z##
##6x+6y+6z=0##
chwala said:
Then, it follows that the two equations are dependent i.e
##6x+6y+6z= 2(3x+3y+3z)##
thus the system cannot have a unique solution.
It seems to have led to the correct result, but it's a little hard to follow your work.

Working with the 1st and 2nd equations, you have:
##4x+5y+6z = 1##
##x+2y+3z = 1##
and you subtracted the 2nd equation from the first to get ##3x + 3y + 3z = 0##. It would have been helpful to indicate that you did that subtraction to get your third equation. BTW, that equation is equivalent to x + y + z = 0.

You did essentially the same thing working with the 2nd and 3rd equations. Geometrically you found that the planes represented by the 1st and 2nd equations intersect in the plane x + y + z = 0. Similarly, you found that the planes represented by the 2nd and 3rd equations also intersect in the plane x + y + z = 0.

Rather than working with equations as you did, IMO it's much better to either calculate the determinant of the matrix of coefficients to show that it's zero, or to use Gauss-Jordan elimination to show that you end up with one or more rows of zeroes.

chwala said:
In my approach from 1.i,
##3x+3y+3z=0##
##x+y+z=0##
##x+y+\dfrac{1}{3} (1-x-2y)=0##
##3x+3y+1-x-2y=0##
##2x+y=-1##
##y=-2x-1##
This is really hard to follow with no explanation of what you're doing. Eqn. 1 above apparently comes from subtracting the 1st given equation from the 2nd given equation.
Eqn. 2 is the simplification of Eqn. 1.
Eqn. 3 apparently comes from solving for z in the 1st given equation and substituting it into Eqn. 2.
Eqns. 4, 5, and 6 come from various simplifications.

Using Gauss-Jordan elimination I find a solution that includes these two points: (-1, 1, 0) and (0, -1, 1) plus an infinite number of other points.
 
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